1037 Magic Coupon （25 分）思维

1037 Magic Coupon （25 分）

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N's!

For example, given a set of coupons { 1 2 4 −1 }, and a set of product values { 7 6 −2 −3 } (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.

Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.

Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons N​C​​, followed by a line with N​C​​ coupon integers. Then the next line contains the number of products N​P​​, followed by a line with N​P​​ product values. Here 1≤N​C​​,N​P​​≤10​5​​, and it is guaranteed that all the numbers will not exceed 2​30​​.

Output Specification:

For each test case, simply print in a line the maximum amount of money you can get back.

Sample Input:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

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代码

#include<bits/stdc++.h> 
using namespace std;
int main(){
	vector<int>a,b;
	int n,x;
	scanf("%d", &n);
	for(int i = 0 ; i < n; i++)
	{
		scanf("%d",&x);
		a.push_back(x);	
	}
	int m;
	scanf("%d", &m);
	for(int i = 0 ; i < m; i++)
	{
		scanf("%d",&x);
		b.push_back(x);	
	}  
	sort(a.begin(),a.end());
	sort(b.begin(),b.end());
	long long sum = 0;
	for(int i = 0; a[i] < 0 && b[i] < 0; i++)
		sum += a[i] * b[i];
	for(int i = n - 1, j = m - 1; a[i] > 0 && b[j] > 0; i--,j --)
		sum += a[i] * b[j];	
	printf("%lld\n",sum); 
  return 0;
}